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In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. `0V,2.0A`B. `0V,1.4A`C. `5.6V,1.4A`D. `8V,2.0A` |
Answer» Correct Answer - C `X_L=omegaL=(5 xx10^(-3))(2000)=10 Omega` `X_C=1/(omegaC)=1/((2000)50xx10^(-6))=10Omega` Since, `X_L=X_C` circuit is in resonance. `Z=R=(6+4)=10Omega` `I_("rms")=V_("rms")/Z=((20//sqrt2))/10=1.414A` This is also the reading of ammeter. `V=4I_("rms")` `~~5.6` Volt. |
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